Percent Yield

  One way to think of percent yield is as a measure of success. To find this number you will need two things: the theoretical yield and the experimental yield. The theoretical yield is the amount of products that should be formed and the experimental yield refers to the amount you're given based on the experiment. Because you are usually given the experimental yield, you will need to find the theoretical yield first...

EX.1:

If you burn 12 g of carbon (C) to produce 48.0 g of carbon dioxide (CO2) what is the percent yield?

1. The first and one of the most important steps is to make sure you have a correct and balanced equation.

C + O₂ --- > CO₂

2. You will then have to find the theoretical yield of carbon dioxide because you are given the experimental value. Take what you have and convert it into moles as it is the most common conversion factor.

12 g of C    x    1 mol of C

                      12.0 g of C

3. Next, use your mole to mole ratio: what you need over what you have. Cancel out what you can.

12 g of C    x    1 mol of C    x    1 mol of CO2

                      12.0 g of C            1 mol of C

4. After having your moles of Carbon Dioxide, use molar mass as your conversion factor and cancel out what you can.

12 g of C    x    1 mol of C    x    1 mol of CO2    x    44.0 g of CO2

                      12.0 g of C            1 mol of C            1 mol of CO2

5. Multiply and divide your equation and you will now have your theoretical yield. Plug it into the percent yield formula.

% YIELD    =    Experimental     x 100

                           Theoretical

% YIELD    =    48.0 g    x    100

                          44.0 g

% YIELD    =    109%

For more examples check out this link:

 http://www.youtube.com/watch?v=TKNxdL7DN1I

 

Mass to Mass Conversions

  In order to complete mass to mass conversions, there is one additional conversion you have to make that involves moles.

EX.1: How many grams of O2 are there when 10 g of Al2O3 decomposes?

1) To start off, create a balanced equation.

Al2O3  ->  Al + O2

2Al2O3  ->  4Al + 3O2

2) Since you are given the mass of Al2O3, use the molar mass of the compound in order to cancel out the grams and left with moles

10 g of Al2O3   x   1 mol Al2O3

                           102.0 g Al2O3

3) Next, you will have to cancel out the moles of Al2O3. One way to look at the co-efficients of a compound is to think of them as moles, which you can use to cancel out Al2O3. Put what you know on the bottom and put what you need on the top.

10 g of Al2O3   x   1 mol Al2O3   x   3 mol O2

                           102.0 g Al2O3      2 mol Al2O3

4) Now that you have the moles of O2, you can now use molar mass as a conversion factor in order to find out what you're looking for which the mass of O2.

10 g of Al2O3   x   1 mol Al2O3   x   3 mol O2    X   32.0 g O2

                           102.0 g Al2O3      2 mol Al2O3     1 mol O2

5) Finally, calculate your answer and leave it in its significant digits.

10 g of Al2O3   x   1 mol Al2O3   x   3 mol O2    X   32.0 g O2   =   4.7 g of O2

                           102.0 g Al2O3      2 mol Al2O3     1 mol O2

For more examples about mass to mass or moles to mass conversions, check out this video:

http://www.youtube.com/watch?v=sTsshxcHhaI

Density conversion:

http://www.youtube.com/watch?v=fqLCwuKMBMA

1. question: Water has a density of 1.0g/ml. Determine the mass of 11.5ml of water

                     How many moles are in 11.5 ml of water?

    answer: 10g/ml x 11.5ml=11.5g=12g

                  11.5g x 1mol/ 18.0g=0.64 mol

2. question: An unknown compound has a molar mass of 65.0g/ mol. If 0.25mol occupies 

                    a volume of 50ml determine the compound's density

    answer: 0.25mol x 65.0g/ mol= 16.25g 

                  16.25/50ml= 0.325g/ml= 0.33g/ml

3. question: The density of aluminum is 2.70g/ml. A solid piece of aluminum has a volume

                     of 45.0ml. Determine the number of aluminum atoms present

    answer: 2.70g/ml x 45.0ml x 1mol/27.0g x 6.02 x 10^23 atoms/ 1 mol

                  = 2.71 x 10^24 atoms

4 question: Copper has a density of 8.96g/ ml. Determine the numbers of atoms in a copper

                   key that has a volume of 20ml TRY THIS YOURSELF!! :D

Density of gases:

1. The density of gases varies with temperature 

2. An STP we can find density by: 

    MM [molar mass]/ 22.4 l/mol [molar volume]

1. question: Calculate the density of 0,2 at STP

     

    answer: 32.0g/mol / 22.4l/mol=1.43g/L 

2. question: Determine the density of methane[ CH,4] at STP

    answer: 16.0g/mol / 22.4L/mol=0.714g/L

Stoichiometry

Hello fellow subscriber... and viewers! Today we learned about the Quantitative Chemistry - Stoichiometry.

....What? What is Stoichiometry? WELL it is a branch of chemistry that deals with the qualitative analysis of chemical reactions.

To understand Stoichiometry, our teacher reviewed us on the 6 types of reactions (not including complex reactions that doesn't go under the 6 common reactions). This is what we learned in grade 10; last year.

Ok, so there are 6 common types of reactions
1) Synthesis (Formation)
2) Decomposition
3) Single Replacement (SR)
4) Double Replacement (DR)
5) Neutralization
6) Combustion

Here is a breakdown of each reaction:
1)Synthesis : Basically a reaction to make compunds; so the equation would be-
A+B=AB

2)Decomposition : The opposite of Sythesis, which breaks down compounds.
AB= A+B

3)Single Replacement: When an element switches with another element in the equation, similar to this:
A+ BC=B+AC

4)Double Replacement: When both elements switch... like this
AB+CD=AD+BC

5) Neutralization : A reaction between an acid and a base, to form a compound and HOH
ACID+BASE= COMPOUND + HOH

6)Combustion: A reaction usually dealing with a hydrocarbon, that ALWAYS produces CO2 and HOH
Hydrocarbon + Oxygen = CO2 + HOH

There we have it, introduction to the world of Stoichiometry! Tune in next time for more!

PERCENT COMPOSITION

-the percentage by mas of an elemet in a compund is always the same
-to find the percent by mass determine the mass of each element present in one mole

http://www.youtube.com/watch?v=CGQoQk-xZ68

Empirical & Molecular Formulas

EMPIRICAL FORMULAS: 

  The simplest formulas of compounds are called Empirical Formulas. They do not show the actual number of atoms but rather the simplest ratios the compound has.

EXAMPLE:

Where Nitrogen can be written as N2 when alone, it's empirical formula would just be N.

Similarly, where Dinitrogen Tetraoxide is written as N2O4, it's empirical formula would just be NO2.

  To determine the empirical formula, we need to know the ratio of each element. For that we will need to prepare a table to help us:

Atom | Mass | Molar Mass | Moles | Smallest Mole | Ratio 

EXAMPLE:

Let's start with 4.58 g of Silicon and 5.21 g of Oxygen.*

*You can find this example and others on: http://homepage.mac.com/sklemmer/Honors/empformulawkst.htm 

1. Create a table.

 Atom | Mass | Molar Mass | Moles | Smallest Mole | Ratio 

2. Fill in the table with what you know. To figure out the number of moles, find the molar mass of each element.

Atom | Mass | Molar Mass |   Moles   | Smallest Mole | Ratio 

Si       4.58 g   28.1 g/mol    0.163 mol

O        5.21 g   16.0 g/mol    0.326 mol

3. To fill in the column for the ratios, you will need to use your smallest mole. What this means is that you take your smallest value of moles which would be 0.163 mol in this case, and divide it from the number of moles of each atom.

Atom | Mass | Molar Mass |   Moles   | Smallest Mole | Ratio

Si       4.58 g   28.1 g/mol    0.163 mol÷0.163= 1          -> 1

O        5.21 g   16.0 g/mol    0.326 mol÷0.163= 2          -> 2

4. Write out the final formula with the ratios in the place of the subscripts. This will be the empirical formula.

SiO2

MOLECULAR FORMULAS:

  While empirical formulas give the simplest ratios, Molecular Formulas give the actual number of atoms. To know the molecular formula, you would need to know the empirical and molar mass.

EXAMPLE:

Empirical   |   Molecular 

C2H6O                ?

     ?              138 g/mol

1. Calculate the molar mass of the Empirical Formula and place it in the table.

Empirical   |   Molecular 

C2H6O                ?

46 g/mol         138 g/mol

2. Figure out the number of times 46 goes into 138 to find out the multiplier.

138÷46= 3

3. Use the multiplier as a co-efficient for the empirical formula and expand. This will give you the molecular formula.

Empirical        |       Molecular 

C2H6O-> 3(C2H6O) -> C6H18O3

46 g/mol               138 g/mol

*For more information on Empirical and Molecular Formulas check out this link:

http://www.youtube.com/watch?v=gfBcM3uvWfs

MOLAR VOLUME LAB: [ russel m., aaron l., and trevor n.]


The task in this lab was to experimentally determine the molar volume of a gas


Materials: Butane, Sink, Water , 100mL Graduated cylinder, Weigh scale


Prelab:
The chemical formula for butane is C4H10
The molar mass for butane is 74g


Analysis: The mass of butane lighter before the experiment was 21.65g
                The mass of the butane lighter after the experiment was 21.40g
                The mass of butane used in the experiment was 0.25g
                There were 0.004 moles of the mass of butane 
                The molar volume of butane is 12.5L/mol